Tài liệu Physics exercises_solution: Chapter 42 ppt

However, the thermal energy at 300 K is much less than the bond energy of H , so we would expect it to remain intact at 2room temperature... 42.7 the spacing between adjacent vibrational

42.1: a)

K)J1038.1(3

)eVJ1060.1)(

eV109.7(23

22

3

23

19 4

)eVJ1060.1()eV48.4(2

temperature will be more than enough to break up He2 However, the thermal energy at

300 K is much less than the bond energy of H , so we would expect it to remain intact at 2room temperature

42.2: a) 5.0eV

4

1 2 0

b) 5.0eV(4.3eV3.5eV)4.2eV

42.3: Let 1 refer to C and 2 to O.

nm1128.0,kg10656.2,

kg10993

0 2 1

m r

)carbon(nm0484.0

0 2 1

m r

b) 2 1.45 10 46kg m2;

2 2

2

1

42.4: The energy of the emitted photon is 1.0110 5eV, and so its frequency and wavelength are

GHz44.2)

sJ1063.6(

)eVJ1060.1)(

eV1001.1(

34

19 5

.m123.0Hz)1044.2(

)sm1000.3(

This frequency corresponds to that given for a microwave oven

42.5: a) From Example 42.2,

2 46

23

1 0.479meV7.67410 JandI 1.44910 kgm

E

srad1003.12

m8.49)srad1003.1)(

m100484.0

)J10083.2(22

gives2

26 2

r

0 max

2 max r

v m E

b) According the Eq 42.7 the spacing between adjacent vibrational energy levels

is twice the ground state energy:

,)2

1(

E E

ω n

E

n n

eV2690.0(

)sJ1063.6(

T

c) The vibrational period is shorter than the rotational period

42.7: a) 2

H Li

H Li 2

m m

m m r

m

I    

eV

1053.7J1020.1

mkg1069.3

)sJ10054.1(4

4))13()3()14(4(2

mkg1069

3

)kg1067.1kg1017.1(

)m1059.1)(

kg1067.1)(

kg1017

1

(

3 21

2 47

2 34

2 2

3 4

2 47

27 26

2 10 27

E E

J1020.1

s)m1000.3)(

sJ1063.6(

21

8 34

E I

l l E I

l l

2 2

2 2 2

1

2

)(

22

)1(,

2

)1

E h

E

f

22

42.11: Energy levels are

I l

l ω n

E E

2)1(2

592

4

)eVJ10602.1)(

eV2700.0(

)sm10998.2)(

sJ10626.6(λ

eV2700.0)eV10395.2)(

4()eV2690.0)(

1(

6

19

8 34

b) n0,l 2n1,l 1:

m10627

4

λ

)eVJ10602.1)(

eV2680.0(

)sm10998.2)(

sJ10626.6(λ

eV2680.0)eV10395.2)(

4()eV2690.0)(

1(

6

19

8 34

c) n0,l 3n1,l2:

m

10634.4)eVJ10602.1)(

eV2676.0(

)sm10998.2)(

sJ10626.6(

λ

eV2676.0)eV10395.2)(

6()eV2690.0)(

1(

6 19

8 34

used

)]

Hz1024.1(2)[

kg1015.3()kg1067.1(

22

26 27

2 14 26

27

2 1

2 2 1 2 r

m m

πf m m πf

m k

3

eV513

0

J1022.8Hz1024.1sJ1063

sm1000.3

42.14: a) As a photon,

eVJ1060.1eV1020.6

sm1000.3sJ1063.6

λ

19 31

λ

19 27

m1049.4

kg105.89kg1082.3

3 3

3 29

26 26

Cl Na

42.16: For an average spacing a, the density is ρm a3, where m is the average of the

ionic masses, and so

3 3

25 26

)mkg1075.2(

2kg1033.1kg1049

)sm1000.3()sJ1063.6(λ

13 13

8 34



So the number of electrons that can be excited to the conduction band is

eV12.1

eV1034

42.19: a) To be detected the photon must have enough energy to bridge the gap width

sm1000.3sJ1063.6

19

8 34

42.20: 3 1.17 105 m s

theorem does not hold for the electrons at the Fermi energy Although these electrons are very energetic, they cannot lose energy, unlike electrons in a free electron gas

42.21: a) Eψ

z

ψ y

ψ x

2 2

2 2

πy n L

πx n A

ψ sin x sin y sin z

π n x

2 2

2 2

2

2

mL

π n n n E

Eψ ψ L

π n L

π n L

π n m

z y x

z y

2

31

mL

π E n

 and the degeneracy is    2  3 6

The second excited state       2 22

2

92

,2,1or2,1,2or1,2,2

πz n dy

L

πy n dx

L

πx n A

1

eVJ1060.1Jstates10

5

9

s)J10054.1(2

)eVJ101.60(eV)5.0)(

m101.0(kg))10

9.11(2(2

2

22

19 40

3 34 2

2 19

2 3

6 2

31

2 3 2 2

g

E π

V m

E

g



42.24: Equation (42.13) may be solved for n  mE 2 L π

rs  2 , and substituting this into Eq (42.12), using L3 V , gives Eq (42.14)

42.25: Eq.(42.13): 2

2 2 2 rs

2mL

π n



.103.4

)eVJ1060.1()eV7.0()kg109.11(2)sJ101.054(

010.02

7 rs

19 31

34 rs

L n

eVJ101.601.942

KJ1038.1

eVJ1060.1eV23

kT π

K300KJ1038.1

23 2

F 2

K

molJ194.00233

KmolJ

c) Mostly ions (see Section 18.4)

42.28: a) See Example 42.10: The probabilities are 1.7810 7,2.3710 6, and

f F 1 (see Problem (42.46)), and so the probability that a state at the top

of the valence band is occupied is the same as the probability that a state of the bottom of the conduction band is filled (this result depends on having the Fermi energy in the middle of the gap)

1lnK300KJ1038.1

1

1ln

20 F

4

23 F

E

E E

E f kT E E

So the Fermi level is 0.20 eV below the conduction band

42.30: a) Solving Eq (42.23) for the voltage as a function of current,

.V0645.01mA3.60

mA40.0ln1

I e

2 19

A1025.9

580 0

h E

mC10

c) q e0.78

d)

0590

C104.9m101.6

mC10

10 30

e q d

p q

42.34: The electrical potential energy is U 5.13eV, and

m

108.24

42.35: a) For maximum separation of Na andCl for stability:

.m106.9)J102.40(4

)C101.60(

J10.402eVJ101.60eV3.6eV5.14

10 19

0

2 19

19 19

0 2

r πε

e U

b) For K andBr:

.m108.1J)1028.1(4

)C101.60(

J10.281eVJ101.60eV3.5eV4.34

9 19

0

2 19

19 19

0 2

r πε

e U

42.36: The energies corresponding to the observed wavelengths are 3.2910 21J,

.J101.65andJ1006.2,J1047.2,

42.37: a) Pr (44.36) yields I 2.7110 47 kgm2, and so

Cl H

Cl H r

)(

m m

m m I m

)kg105.81kg101.67(

)kg105.81kg101.67()mkg102.71(

10

26 27

26 27

2 47

1

2 2

I

l l l l

l I E l

hc E

m1084.4:m1004.6

;5:m1064.9λ

6:m1004.8λ

;7:m1090.6λ

4 5

5 5

l l

c) The longest wavelength means the least transition energy (l 1l 0)

m

1085.4λ

J1010.4mkg1071.2

s)J10054.1()1(

4

22 2

47

2 34

d) If the hydrogen atom is replaced by deuterium, then the reduced mass changes

m

188)95.1()m4.96(λ:45

m

156)95.1()m4.80(λ:56

m

134)95.1()m0.69(λ:67

m

118)95.1()m4.60(λ:78

forSo

λ)95.1(λkg01.62

kg1016.3λλ

22

λλ

27 27

r r

2 2

μ μ

l l

μ μ

l l

μ μ

l l

μ μ

l l

μ μ

l l

m m

l

r m πc l

I πc hc

42.38: From the result of Problem (42.9), the moment inertia of the molecule is

2 46

2

2

mkg1043.64

hl E

r

m I r

42.39: a)

2

)1(2

2 2

I

l l I

L

E ex   

),0(

E g

and there is an additional multiplicative factor of 2l + 1 because for each l state there are

really2l 1 m l states with the same energy

0

2

)12

b) T 300K,I 1.44910 46kgm2

)mkg10449.1(2

)11()1

2 46

J1067.7

)12()2(

23 2

5(

.5)1

2

(

0556 0 0

2

)110()10(

23 2

2

)120()20(

23 2

41(

.41

42.40: a) 1.44910 46 kgm2.

co

I

.0E

J

1067.7)mkg10449.1(2

)11()1()sJ10054.12

)1(

0

23 2

46

2 34 2

.eV1079.4J1067

J1067.7(

s)m1000.3()sJ1063.6(

23

8 34

(a))

part(fromJ1067.7

J

102.76K)(20K)J1038.1(

23

22 23

kT

Therefore, although T is quite small, there is still plenty of energy to excite CO molecules

into the first rotational level This allows astronomers to detect the 2.59 mm wavelength radiation from such molecular clouds

42.41: a) 2

Cl Na

Cl Na 2

m m

m m r

m I







)kg108068.5kg108176.3(

)m10361.2()kg108068.5()kg108176.3(

26 26

2 10 26

01148.0λ

J10730.1mkg10284.1

s)J10054.1(22)26

2 45

2 34 2

1 2

I E

0

0 1

E E l

cm

297.2λ

J10650

b) Carrying out exactly the same calculation for Na37Cl, where mr(37) 3542 kg

10 26 and I(37)1.31210 45kgm2 we find for

cm

347.2λandJ10465.8:

01

Forcm

173.1λandJ10693.1:

12

l

E l

l

So the differences in wavelength are:

cm

050.0cm2.297cm

347.2λ:01

cm

025.0cm1.148cm

173.1λ:12

l l

42.42: The vibration frequency is, from Eq (42.8),   1.121014

2( 2 r 

22

12

1

m

k E

m

k n

)mN2(576)

sJ10054.1(2

42.44: a) The frequency is proportional to the reciprocal of the square root of the

reduced mass, and in terms of the atomic masses, the frequency of the isotope with the deuterium atom is

.)(

1

)(

1)

(

)

H F

D F 0

2

F D D F

F H H F

0      



m m

m m f

m m m m

m m m m f f

Using f0from Exercise (42.13) and the given masses, f 8.991013Hz

I H

2 I H 2

r m m r m I

2

1024

r 2

2:since

2

12

)1(

2

12

)1(

m

k πf ω hf

n I l l

m

k h

I l l

.1093.6Hz1096.3

sm1000.3

2λ

.2

12

32)02(

6

13 11

8 2

2 2

I

c hf

I

hc E

hc

hf I

hf I

sm1000.3

22λ

2)26(

6 13

11

8 2

hf I E





iii) n2,l 2n1,l 3

.m1040.4Hz1093.6)Hz1096.3(3

sm1000.3

23λ

2)126(

6 13

11

8 2

hf I E







42.46: The sum of the probabilities is

.1

111

1

11

1)

()(

/

F F

kT E kT

E

kT E kT

E

e

e e

e e

E E f E E f

42.47: Since potassium is a metal we approximate EF EF0

.eV2.03J

1024.3)

kg1011.9(2

)/m1031.1(s)J10054.1(3

melectron10

31.1kg10

49

6

mkg

851

ionconcentratelectron

the

But

23

19 31

3 3 28 2

34 3

3

F

3 28

26 3

3 2 3 3

n π

42.48: a) First we calculate the number-density of neutrons from the given mass-density:

.m102.4)neutronkg

1067.1/)mkg100.7

kg1067.1(2

)m102.4()2sJ1063.6(32

27

3 44 2

34 2

0

F

3 3

3 3 3 3

b) Set kT EF0(see Exercise 42.26) to obtain

.K103.1)KJ1038.1(

)J108.1

28

11 0

42.49: a) Each unit cell has one atom at its center and 8 atoms at its corners that are each

shared by 8 other unit cells So there are 18 82atoms per unit cell

3 28

3

9 4.66 10 atoms m)

m1035.0(



V n

b)

3 2

3 3 F0

π

In this equation N V is the number of free electrons per m3 But the problem says to

assume one free electron per atom, so this is the same as V n calculated in part (a)

kg10109

Setting this equal to zero when r  givesr0

2 0 7

0

48

αe

πε A

r and so

.8

1

7 0 0

αe U

At r r0,

.eV7.85J

1026.132

0 0

2

r πε

αe U

b) To remove a NaClion pair from the crystal requires 7.85 eV.When neutral

Na and Cl atoms are formed from the Na and  Cl atoms there is a net release of energy

eV,1.53eV

35

3

tot

3 2

3 3 0

V

N m

π E

3

.5

3

2

35

332

3 2

3 3 tot

3 2

3 3

2

3 2

3 3 tot

π dV

dE P

V

N m

π

V

N V

N m

π N

atm1076.3Pa1080.3

)m1045.8()kg1011.9(5

)sJ10054.1(3

5 10

3 3 28 31

2 34 3

c) There is a large attractive force on the electrons by the copper ions

42.52: a) From Problem (42.51):

.35

5

335

.5

3

2

3 2

3 3

3 2

3 3

p

V

N V

N m

π V

dV

dp V B

V

N m

π p

33

Pa106.33

2 2 2

0

F

3000

23

100

23

)100(

2100

c m π

c m V

N mc

E

.m

1.67

m10

3 33

3 28

c) The number of electrons is 6.03 10

kg101.99

kg)10

1003

3 6 3

N e

d) Comparing this to the result from part (a) 400

m101.67

m106.66

3 33

3 35

42.54: a) Following the hint,

dr r

e π r

e π d dr

k

r r

3 0

2

0 2

2

14

1

3 0

k ω

q q π

d r d r

d

11

1111

2 3

0

2 3

2 2

2

4

24

21

4

2

221

1

1

d πε

p r

πε

p r

d d πε

q

U

r

d r r

d r

d r

r

q q πε

U

0

41

.4

24

2

14

22

2224

1111

114

3 0

2 3

0 2

3 2

0

2 3

2

0 2 0 2

r πε

p d

πε

p U

r

d d πε

q r

d r r d πε q

d r d r d r r d πε q

2

r πε

2

r πε p

U  